Find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. A quantity of...

Work is done when a constant force F is applied to move an object a distance D. It is defined with a formula `W = FD.`

For expanding gas, we denote the work done as `W =P* DeltaV.`

With the stated assumption pressure is inversely proportional to volume, we let `P =k/V ` where k is the proportionality constant.

Then plug-in `P = k/V` on `W =P* DeltaV` , we get: `W =k/V* DeltaV `...

Work is done when a constant force F is applied to move an object a distance D. It is defined with a formula `W = FD.`

For expanding gas, we denote the work done as `W =P* DeltaV.`

With the stated assumption pressure is inversely proportional to volume, we let `P =k/V ` where k is the proportionality constant.

Then plug-in `P = k/V` on `W =P* DeltaV` , we get: `W =k/V* DeltaV ` or `(kDeltaV)/V`

The integral of work done  will be `W=int_(V_1)^(V_2)(kdV)/V`

To solve for the proportionality constant `(k)` , we plug-in the initial condition:

`P= 2500` pounds and `V_1= 1` on `P= k/V.`

`2500 = k/1`

`k = 2500*1 =2500`

To solve for the work done by the gas to expand the volume, we plug-in `k=2500` , `V_1=1` , and `V_2=3` on `W=int_(V_1)^(V_2)(kdV)/V` .

`W=int_(1)^(3)(2500(dV))/V`

Apply basic integration property: `int cf(x)dx = c int f(x)dx` .

`W=2500int_(1)^(3)(dV)/V` .

Apply basis integration formula for logarithm.

`W=2500 ln(V)|_(1)^(3)`

Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`W=2500 ln(3)-2500 ln(1)`

`W=2500 [ ln(3)- ln(1)]`

Apply natural logarithm property: `ln(x/y) = ln(x)- ln(y)` .

`W=2500 [ ln(3/1)]`

`W=2500 ln(3)` or `2746.53` ft-lbs.